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Q: Prove that if $X$ is a limit point compact and Hausdorff space and $f$ is a continuous map sends $X$ to real numbers space, then the image of $f$ is compact subspace of $\mathbb{R}$?

I know that the image of limit point compact is not limit point compact in general, hence it is not compact. So the Hausdorff condition plays big role here.

To prove compactness it is sufficient to prove it is closed and bounded. I prove it is closed but I got stuck in boundness, so how can I prove that it is bounded here. Or is there any another way to prove it?

algae
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Ahmed
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2 Answers2

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Suppose the image is not bounded. Then we can find a sequence $(x_n)$ such that $|f(x_n)| \to \infty$. Let $x$ be a limit point of this sequence. Consider $f^{-1}[(f(x)-1,f(x)+1)]$. This is an open set containing $x$ and hence it must contain $x_n$ for infinitely many values of $n$. But this means $|f(x_n)-f(x)| <1$ infinitely many values of $n$. Now use the fact that $|f(x_n)| \leq |f(x_n)-f(x)|+|f(x)|$ to get a contradiction.

Henno Brandsma
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Kavi Rama Murthy
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$X$ is limit point compact and $T_1$ so $X$ is countably compact (e.g. see here if you don't know that) and $f[X]$ is countably compact (this property is preserved by continuous maps, the proof is analogous to the compact case). In $\Bbb R$, being metrisable, countably compact and compact are equivalent. So $f[X]$ is compact.

Henno Brandsma
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